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Ulrich MansmannIBE, Medical School, University of Munich

Group testing 2

Content of the lecture

• How to assess the relevance of groups of genes:- outstanding gene expression in a specific group compared

to other genes;- differential gene expression not of single genes but over a

specific group of genes.

• How to define gene-groups:- exploratory research produces functional groups and genomic

signatures: confirm the relevance of the specific group.- Bioinformatic algorithms can be used to define pathways and

functional groups

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Example I: Cyclin D1 Action

• Lamb J et al. (2003) A mechanism of Cyclin D1 Action Encoded inthe Patterns of Gene Expression in Human Cancer, Cell, 114: 323-334

• Cyclin D1 expression signature: cyclin D1 target gene set.

• Cyclin D1 activity in Human Tumors: Does the cyclin D1 targetgene set play a prominent role in different tumor entities?Being present as highly expressed genes.

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The ideas behind the analysis

Problem:

Two groups of genes have to be compared with respect to gene expression: Is the gene expression in gene group A different from the expression in gene group B. Important: Genes in both groups are different!

Basic idea:

nA genes in group A, nB genes in group B

Order the genes with respect to the expression value. If there is a difference in expression level between both groups, the expression values will be separated. The position of a value in group A will have the tendency to be in general high or low. In case of no difference, the values will be nicely mixed.

Group A

Group BRank of expression value

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The ideas behind the analysis

Group A (nA=10)

Group B (nB=5)

-5 -5 -5 -5 -5 -5 -5 -5 -510

Genes ordered by rank of expression

10 10 10 10-5

MinimumIs the minimum extreme with respect to random group mixing?

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The algorithm fomalized

Basic idea:

• nA genes in group A, nB genes in group B. • Order the genes with respect to expression values.

• Create a vector vv of (nA+nB) components with value – nB at each position where avalue from group A is sitting and with value nA at each position where a value fromgroup B is sitting.

• Calculate yy = cumsum(vv).

• Draw a line starting at (0,0) through points (i, yy[i]). The line will end in (nA+nB,0)because (-nB)⋅nA+ nA⋅nB =0.

• Look at Mvv = max{|min(yy)|,max(yy)} which will be large in case of a good separationbetween both groups.

• Permute the vector vv to get vv*, calculate yy* and Mvv*. Use permutation to calculatethe distribution of Mvv under the Null hypothesis, determine the permutation based p-value: pperm = #{Mvv* ≥ Mvv}/ # permutations.

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Example II: Colon Cancer

Study: 18 patients with UICC II colon cancer, 18 patients with UICC III colon cancer, HG-U133A, 22.283 probesets representing ~18.000 genes. Snap-frozen material, laser microdisection.

Question 1: Are there specific cancer related pathways with a more distinct differential gene expression between UICC II/III?

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Gene set enrichment – Colon cancer

1407 probe sets are studied which belong to 9 cancer specificpathways.

androgen_receptor_signalling 122 apoptosis 245cell_cycle_control 51 notch_delta_signalling 50 p53_signalling 45ras_signalling 316tgf_beta_signalling 100tight_junction_signalling 425wnt_signalling 214

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Gene set enrichment – Colon cancer

group.A group.B Myy p.value

androgen_receptor_signaling 118 1289 6983 0.0568

Apoptosis 238 1169 17801 0.7438

cell_cycle_control 51 1356 10413 0.3616

notch_delta_signalling 50 1357 9010 0.6492

p53_signalling 45 1362 12390 0.0924

ras_signalling 311 1096 15486 0.6252

tgf_beta_signaling 100 1307 22615 0.0128

tight_junction_signaling 406 1001 15456 0.4414

wnt_signaling 214 1193 16318 0.8432

Restriction of the analysis to genes in cancer specific pathways

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Example III: Lymph node metastases

Bertucci F et al. (2004) Gene expression profiling of colon cancer by DNA microarrays and correlation withhistoclinical parameters, Oncogene 23, 1377–1391

Bertucci at al present a gene signature consisting of 46 genes which is claimed to be able to discriminate between LN- and LN+ colorectal cancer.

Is it possible to prove with new data that the signature has discriminative value. Can we reject the Nullhypothesis

P[Y|X] = P[Y].

Y : LN+/LN-X: expression pattern of 46 genes

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Goeman’s Global Test

• Test if global expression pattern of a group of genes is significantlyrelated to some outcome of interest (groups, continuous phenotype).

• If this relationship exists, then the knowledge of gene expressionhelps to improve the prediction of the phenotype of interest. If theprediction can not improved by knowing the gene expression thenthere will not be differential gene expression.

• Test statistic:Q ~ (Y-µ)’R (Y-µ)

~ Σ [Xi’(Y-µ)]² sum over genes of the pathway~ Σ Σ Rij(Yi-µ) (Yj-µ) sum over subjects

µ: Mean of phenotype, Xmi Expression for gene m in subject iR : X’X matrix of correlations between gene expression of subjects

Goeman JJ. Et al. (2003) A global test for groups of genes: Testing association with a clinical outcome, Bioinformatics, 20:93-99; Bioconductor package: globaltest

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Example III: Lymph node metastases

Test is not significant (p=0.43)

No clear answer on the predictive power of the signature.

No evidence for a difference is not evidence for no difference!

Question of power

Reasons for a non-significance: bad experiment or ...?

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Example II: Colon Cancer

Study: 18 patients with UICC II colon cancer, 18 patients with UICC III colon cancer, HG-U133A, 22.283 probesets representing ~18.000 genes. Snap-frozen material, laser microdisection.

Question 2: Is there differential gene expression in the p53 signalling pathway between UICC II and UICC III colon cancer?

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Goeman’s Global Test – Example II

• Test for differential gene expression in p53 signalling pathway45 probesets

• Global Test result:45 out of 45 genes used; 36 samples

p value = 0.0114 based on 10000 permutations

Test statistic Q = 11.78 with expectation EQ = 5.466and standard deviation sdQ = 2.152 under the null hypothesis

• Informative plots:Sample plot: how good fits a sample to its phenotypeCheckerboard: Correlation between samplesGene plot: Influence of single genes to test statistics

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Goeman’s Global Test – Example II

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Σ [Xi’(Y-µ)]²Σ Σ Rij(Yi-µ) (Yj-µ)

Values dichotomized around median

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Goem

an’sG

lobal Test –E

xample

II

-50 0 50 100

influence

Co1.T.IT.30.F.CEL

Co14.T.IT.88.F.CEL

Co15.T.IT.558.F.CEL

Co20.T.IT.570.F.CEL

Co37.T.IT.669.F.CEL

Co39.T.IT.673.F.CEL

Co40.T.IT.676.F.CEL

Co41.T.IT.680.F.CEL

Co45.T.IT.688.F.CEL

Co581.T.IT.1085.F.CEL

Co612.T.IT.1423.F.CEL

Co618.T.IT.1449.F.CEL

Co619.T.IT.1515.F.CEL

Co659.T.IT.1742.F.CEL

Co670.T.IT.1821.F.CEL

Co672.T.IT.1841.F.CEL

Co675.T.IT.1834.F.CEL

Co9.T.IT.1728.F.CEL

Co10.T.IT.83.F.CEL

Co13.T.IT.86.F.CEL

Co17.T.IT.563.F.CEL

Co23.T.IT.1052.F.CEL

Co44.T.IT.1162.F.CEL

Co5.T.IT.62.F.CEL

Co54.T.IT.1480.F.CEL

Co610.T.IT.1206.F.CEL

Co611.T.IT.1482.F.CEL

Co620.T.IT.1570.F.CEL

Co652.T.IT.1762.F.CEL

Co653.T.IT.1731.F.CEL

Co656.T.IT.1740.F.CEL

Co657.T.IT.1744.F.CEL

Co665.T.IT.1820.F.CEL

Co666.T.IT.1777.F.CEL

Co668.T.IT.1791.F.CEL

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1 samples

0 samples

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Two questions about groups of genes

Question 1: Two groups of genes have to be compared with respectto gene expression: Is the gene expression in gene group A different from the expression in gene group B.

Question 2: Is there differential gene expression between differentbiological entities not in terms of single genes but withrespect to a defined group of genes.

Genes of group A Genes of group B

Entity I Entity II

Well definedgroup of genes